Question

A big box at ${100}^{o}C$ is connected to an ice cube through identical metal rods as shown. Ice is meting at the rate of $Q_2$ gram/sec. Rods are rearranged so that they are connected in series between box and Ice cube. The new rate of melting is $Q_2$ gram/sec. Then $\frac{Q_2}{Q_1}$ ?

• A. 4
• B. 2
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A 4

When in parallel.

$R_{Th}eq=\frac{R}{2}$

When in series;-

$\begin{array}{c}{R}_{Th}eq=2R\\ \therefore \frac{Q_2}{Q_1}=4\end{array}$

Therefore,

Option $A$ is correct.